Use the worked example above to help you solve this problem. A ball is thrown fr

Question

Use the worked example above to help you solve this problem. A ball is thrown from the top of a building with an initial velocity of 21.0 m/s straight upward, at an initial height of 54.9 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure.

1. Determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant.

2. Determine the time needed for the ball to reach the ground.

3. Determine the velocity and position of the ball at t = 5.75 s.

Explanation / Answer

(1)

Time needed for the ball to return to the height from which it was thrown,

t = u / g

t = 21 / 9.8

t = 2.14 s

velocity of the ball at that instant = v = 21 m/s

(2)

time needed for the ball to reach the maximum height = 2.14 s

Maximum height = u^2 / 2g

hmax = 21^2 / 2*9.8 = 22.5 m

time needed to fall from maximum height, t = sqrt (2H / g)

t = sqrt [2*(22.5 + 54.9) / 9.8] = 3.97 s

Total time needed for the ball to reach the ground

T = 2.14 + 3.97

T = 6.12 s

(3)

velocity of ball at t = 5.75 s

time after reaching maximum height,

t = 5.75 - 2.14 = 3.61 s

v = vo + at

v = 0 + 9.8*3.61

v = 35.37 m/s

Position from maximum height,

s = ut + (1/2)at^2

s = 0 + (1/2)*9.8*(3.61)^2 = 63.85 m

position of ball at 5.75 s,

d = (54.9 + 22.5) - 63.85

d = 13.55 m

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